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The equation for the circle is: x^2+y^2−12x+6y−19=0 . What is the center of the circle? Question 12 options: (-6, 3) (12, -6) (6, -3) (-12, 6)

The equation for the circle is: x^2+y^2−12x+6y−19=0 . What is the center of the circle? Question 12 options: (-6, 3) (12, -6) (6, -3) (-12, 6)
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The equation for the circle is:

x^2+y^2−12x+6y−19=0 .



What is the center of the circle?

Question 12 options:

(-6, 3)


(12, -6)


(6, -3)


(-12, 6)

asked
User Andreak
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1 Answer

2 votes
The answer would be ( -12, 6 ) try plugging in answers into the problem
answered
User Bhagyesh Dudhediya
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